four 0.2 kg masses are held together by light rods in the shape of a square, 0.22 m on a side. calculate the…

Explanation:

Step1: Recall the formula for moment of inertia

The moment of inertia ( I ) of a point mass ( m ) at a distance ( r ) from the axis of rotation is given by ( I = mr^2 ). For a system of point masses, the total moment of inertia is the sum of the moment of inertia of each mass, i.e., ( I_{\text{total}}=\sum_{i = 1}^{n}m_ir_i^2 ).

Step2: Analyze the positions of the masses

We have four masses, each of mass ( m = 0.2\space kg ), arranged in a square of side length ( a=0.22\space m ). Let's consider the corner about which we are calculating the moment of inertia.

• Two of the masses are at the corner itself, so their distance from the axis (( r )) is ( 0 ).
• One mass is at a distance equal to the side length of the square (( r = a )) from the corner.
• The fourth mass is at a distance equal to the length of the diagonal of the square from the corner. The length of the diagonal ( d ) of a square with side ( a ) is ( d=\sqrt{a^2 + a^2}=\sqrt{2}a ).

Step3: Calculate the moment of inertia for each mass

• For the two masses at the corner (( r = 0 )): ( I_1 = m(0)^2=0 ) and ( I_2 = m(0)^2 = 0 ).
• For the mass at distance ( a ) from the corner: ( I_3=m a^2 ). Substituting ( m = 0.2\space kg ) and ( a = 0.22\space m ), we get ( I_3=0.2\times(0.22)^2 ).
• For the mass at distance ( \sqrt{2}a ) from the corner: ( I_4=m(\sqrt{2}a)^2=m\times2a^2 ). Substituting the values, ( I_4 = 0.2\times2\times(0.22)^2 ).

Step4: Calculate the total moment of inertia

The total moment of inertia ( I_{\text{total}}=I_1 + I_2+I_3 + I_4 ). Since ( I_1 = I_2 = 0 ), we have: [ \begin{align*} I_{\text{total}}&=0 + 0+0.2\times(0.22)^2+0.2\times2\times(0.22)^2\ &=0.2\times(0.22)^2\times(1 + 2)\ &=0.2\times(0.22)^2\times3 \end{align*} ] First, calculate ( (0.22)^2=0.0484 ). Then, ( 0.2\times0.0484 = 0.00968 ). Then, ( 0.00968\times3=0.02904\space kg\cdot m^2 ).

Answer:

( 0.02904 ) ( kg\cdot m^2 )